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I tried to use the definition and arrived this far:

$|f(x)-f(y)|=\left|\frac{x}{1+|x|}-\frac{y}{1+|y|}\right|=\frac{|x-y+x|y|-y|x||}{(1+|x|)(1+|y|)}\leq|x-y+x|y|-y|x||$.

Any suggestion for ending the proof?

I also tried to prove that $\frac{x}{1+x}$ is uniformly continuous on $[0,\infty[$ and that $\frac{x}{1-x}$ is uniformly continuous on $[-\infty,0[$, but I wonder if we can use the definition with the function $f(x)=\frac{x}{1+|x|}$ itself.

coolsv
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2 Answers2

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You have \begin{align} |x-y+x|y|-y|x|\,|&=|x-y + x (|y|-|x|) +(x-y)|x|\,|\\ \ \\ &\leq|x-y|+|x|\,||y|-|x||+|x|\,|x-y|\\ \ \\ &\leq |x-y|+|x|\,|y-x|+|x|\,|x-y|\\ \ \\ &=(1+2|x|)\,|x-y|, \end{align} which hints that you cannot kill the denominators as you did, because you will need them. If you keep the denominators, you need to bound $$ \frac{(1+2|x|)\,|x-y|}{(1+|x|)(1+|y|)}≤ \frac{(2+2|x|)\,|x-y|}{1+|x|}=2\,|x-y|. $$

Aman Kushwaha
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Martin Argerami
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Here is another approach: Your function $f$ is differeentiable on all of ${\mathbb R}$, whereby $$f'(x)={1\over\bigl(1+|x|\bigr)^2}\qquad(-\infty<x<\infty)\ .$$ As $|f'(x)|\leq1$ for all $x$ the MVT implies that $|f(x)-f(y)|\leq|x-y|$; hence $f$ is even Lipschitz continuous on ${\mathbb R}$.

Christian Blatter
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  • Thank you, too! But the purpose was to use the definition and not the Lipschitz condition ;-) – coolsv May 15 '16 at 20:06
  • The "definition" would talk about $\epsilon$ and $\delta$. Your attempt as well as Martin Argerami's answer are (rightfully) in terms of Lipschitz continuity, and so is my version. – Christian Blatter May 16 '16 at 07:54